is a bijective ring homomorphism 
. The automorphisms of 
form a group, Aut
, under composition. If 
, we define 
to be the subset
1998/99 GCW
An automorphism of a field is a bijective ring homomorphism . The automorphisms of form a group, Aut, under composition. If , we define to be the subset
of . It is easily verified to be a subfield of . If 
is an extension of fields we define 
to be the subgroup
of 
. We call elements of 
-automorphisms of 
. We call the Galois group of the extension.
Theorem: The automorphisms of a field are linearly independent as linear maps of the field.
Proof: Let be a field and let 
be distinct automorphisms of . We have to show that if 
are elements of 
such that
for all 
then 
. We argue by contradiction. Suppose that 
is the least number of the 
's that are nonzero. Up to a relabelling we can suppose that 
are nonzero. Since 
and 
are distinct there is a 
in 
such that 
. Replacing 
in the equation above by 
and 
by 
we get
using the fact that automorphisms preserve multiplication. But multiplying all the way through by 
gives
Subtracting, the last terms cancel, and we get
which gives an equation with a positive number of nonzero coefficients that is less than 
. This contradiction establishes the result.
Theorem: Let be a finite extension of degree 
. Then cannot have more than 
elements.
Proof: Again we argue by contradiction. Suppose that 
are distinct.
Let 
be a basis of as a -vector space. Consider the 
matrix
of elements of . By elementary linear algebra there has to exist a nonzero vector 
such that
But since every element of is a linear combination of the 
's with coefficients in , and the 
's fix elements of we have the contradiction that 
are linearly dependent as - linear maps 
.
Let us describe the quite general notion of a Galois connexion. We suppose thar we have two sets, 
and 
, and a relationship 
for elements 
of 
and elements 
of 
(i.e. a subset of 
). Given 
and 
we define
It should be clear that
We deduce that the 
and 
operations reverse inclusions of subsets, i.e.
and that we have the identities
We call 
a closure operator on subsets of 
. We call 
the closure of 
with respect to 
. We say that 
is closed if 
. We have that 
is a closure operator on subsets of 
. The operators 
and 
establish an inclusion reversing bijection between the closed subsets of 
and the closed subsets of 
.
Let be an extension of fields. Let 
. We have a relationship
between elements 
and elements 
defined by 
. This establishes a Galois connexion between subsets of 
and subsets of 
. If 
is an intermediate extension then 
is the subgroup 
. If 
is a subgroup of then 
is the subfield 
of 
. We know that we have a bijection between the closed subgroups and the closed intermediate extensions. But which are they?
A Galois extension is a finite normal separable extension. The main theorem of Galois theory asserts that if is a Galois extension then every subgroup of and every intermediate extension of is closed for the Galois connexion described above. Furthermore, normal subgroups 
of correspond to intermediate extensions for which 
is a normal extension, and 
may be identified with the quotient group 
.
We can understand a bit better what is going on by considering a simple extension 
of degree 
. So suppose that 
is the minimal polynomial of 
over . Let
be a -automorphism. Since every element of 
can be expressed as a polynomial over in 
, it follows that 
is completely determined by the value of 
. Since 
, applying 
we get 
. So 
has to be a conjugate of 
. The elements of correspond bijectively with the conjugates of 
that are in 
. This means that for a finite separable extension the order of equals the degree 
precisely when the extension is normal.
This gives
Theorem: For a finite separable extension the following are equivalent:
is normal,
Proof: We have already seen that 1) is equivalent to 2). From the general facts about Galois connexions we have
If is an intermediate extension and 
is normal then 
is normal. So
is normal and therefore
By the multiplication theorem we deduce that
which means that 1) 
3).
Suppose that 
and that 
has minimal polynomial 
over of degree 
, and that 
of 
's 
conjugates lie in 
. Write them as 
and let
Then 
has coefficients in 
and has a as a root. So 3) implies that 
has coefficients in 
, and since 
must divide 
we deduce that 
so that 3) implies 1).
Theorem: If 
is a Galois extension and 
is a subgroup then
Proof: By theorems above we have 
. But since 
is normal, so is 
, and so 
. We can write
for some primitive element 
. Consider the polynomial
It has coefficients in 
. Since 
it follows that 
is divisible by the minimal polynomial of 
over 
. Hence we get the reverse inequality 
. This means that 
, which is a subgroup of 
, has order equal to the whole group, so the two are the same.
We have now established that for a Galois extension 
the correspondences
give a bijection between intermediate extensions and subgroups 
of the Galois group.
We say that two intermediate extensions 
are conjugate if there is a 
-automorphism 
such that 
.
Proposition: Two intermediate extensions are isomorphic if and only if they are conjugate.
Proof: One way round (conjugate 
isomorphic) is trivial. For the reverse implication we have to show that any -isomorphism 
extends to a -automorphism of . Let 
where 
for some 
. Let 
be the conjugates of 
over 
. Consider the polynomial
It has coefficients in 
and 
. So, applying 
we get 
. Hence 
for some conjugate 
of 
over 
. If we take 
to be given by
then we see that 
.
It follows from this that intermediate extensions are conjugate if and only if the corresponding subgroups of the Galois group are conjugate.
Given an intermediate extension of a Galois extension it is always the case that 
is a Galois extension. However, 
is not a Galois extension unless it is normal, and this happens precisely when it equals all its conjugates. In that case 
is a normal subgroup of and restriction of -automorphisms from to gives a surjective homomorphism of groups 
whose kernel is 
.